Problem 210 - Calculus Volume 1

Question

[T] An anchor drags behind a boat according to the function \(y=24 e^{-x / 2}-24,\) where \(y\) represents the depth beneath the boat and \(x\) is the horizontal distance of the anchor from the back of the boat. If the anchor is \(23 \mathrm{ft}\) below the boat, how much rope do you have to pull to reach the anchor? Round your answer to three decimal places.

Step-by-Step Solution

Answer

You need to pull approximately 7.384 feet of rope.

Step 1: Understanding the Problem

We have been given the function \( y = 24e^{-x/2} - 24 \) which represents the depth \( y \) of an anchor as it trails behind a boat, depending on the horizontal distance \( x \) from the back of the boat. We need to find \( x \) such that \( y \) is 23 feet below the boat's surface.

Step 2: Set Up the Equation

Set \( y = -23 \) in the given exponential equation to find the corresponding \( x \):\[ y = 24e^{-x/2} - 24 = -23 \]

Step 3: Solve for the Exponential Term

Start by isolating the exponential part \( 24e^{-x/2} \):\[ -23 + 24 = 24e^{-x/2} \]\[ 1 = 24e^{-x/2} \]

Step 4: Divide to Isolate the Exponent

Divide both sides by 24 to solve for \( e^{-x/2} \):\[ e^{-x/2} = \frac{1}{24} \]

Step 5: Use the Natural Logarithm

Take the natural logarithm on both sides to solve for \( -\frac{x}{2} \):\[ \ln \left( e^{-x/2} \right) = \ln \left( \frac{1}{24} \right) \]So, \[ -\frac{x}{2} = \ln (\frac{1}{24}) \]

Step 6: Solve for x

Multiply by -2 on both sides to find \( x \):\[ x = -2 \ln \left( \frac{1}{24} \right) \]Compute \( x \) using a calculator to ensure accuracy.

Step 7: Perform Calculation

Using a calculator, compute:\[ x = -2 \times \ln \left( \frac{1}{24} \right) \]\[ \approx 7.384 \] ft (rounded to three decimal places).

Step 8: Conclusion

The horizontal distance from the back of the boat to the anchor, i.e., the amount of rope you have to pull, is approximately 7.384 feet.

Key Concepts

Logarithmic Equations

Logarithmic equations are a fundamental part of algebra that involve logarithms, which are the inverse operations of exponentiation. In simpler terms, if you have an equation like \( b^x = y \), the logarithm helps us find \( x \) by rearranging this relationship into \( x = \log_b(y) \). These equations are useful in scenarios where we know either the base and result, or the inverse is needed to find the exponential function.

In the context of our problem, we used the natural logarithm, \( \ln \), to solve the equation \( e^{-x/2} = \frac{1}{24} \). By applying the natural logarithm to both sides, we could simplify the equation, effectively allowing us to find the value of \( x \). This process of taking the logarithm is crucial when dealing with exponential functions because it simplifies the equation, making it easier to isolate the variable.

This process can be summarized in a few easy steps:

• Identify the exponential equation.
• Apply the logarithm to simplify.
• Use algebraic techniques to solve for the variable.

Mastering logarithmic equations is a powerful tool that aids in solving complex exponential functions efficiently.

Depth Calculation

Depth calculation often involves understanding how a particular parameter changes with another, often influenced by factors like gravity or environmental resistance. In this problem, depth calculation is associated with the depth of an anchor beneath the water surface relative to its horizontal distance behind a boat. This is presented through an exponential function: \( y = 24e^{-x/2} - 24 \).

Solving the problem required setting the equation equal to the given depth, \( y = -23 \) feet, and solving for \( x \). This step represents finding that specific horizontal distance when the depth condition is met.

The steps for depth calculation include:

• Understand the relationship between depth and distance.
• Use the given function properly by substituting known values.
• Solve for the unknown using algebraic techniques.

Effectively understanding depth calculation allows you to apply it in various real-world circumstances where changes in depth depending on distance need to be determined.

Horizontal Distance

Horizontal distance is a crucial variable in problems involving positional changes over a plane or along a surface. In this exercise, determining the horizontal distance was key to understanding how much rope was required to be pulled to retrieve the anchor. The distance \( x \) is linked directly to the function that defines the depth below the boat: \( y = 24e^{-x/2} - 24 \).

To determine this distance, we solved for \( x \) in terms of a known depth, effectively allowing us to understand how far the anchor is in horizontal terms relative to the boat. Using exponential functions and logarithmic equations simplified this process.

Finding horizontal distance is achieved by:

• Using the mathematical relationship relating depth to distance.
• Solving the equation for the horizontal variable.
• Applying calculated results to practical tasks, such as rope retrieval.

Understanding how to compute horizontal distances gives practical solutions to issues like finding the exact fit of objects behind one another or calculating required lengths for tasks.